Q.61 Velocity of an object fired directly in upward direction is given by
V = 80 − 32t, where t (time) is in seconds.
When will the velocity be between 32 m/sec and 64 m/sec?
The velocity equation V = 80 – 32t models an object fired upward under gravity, where deceleration occurs at 32 units per second (likely feet/sec2). The task requires finding the time interval when velocity stays between 32 and 64 units, with the correct answer being option (B) (1/2, 1).
✅ Correct Answer: Option (B) (1/2, 1)
Solve the inequality: 32 ≤ 80 – 32t ≤ 64 for t ≥ 0
Subtract 80: -48 ≤ -32t ≤ -16
Divide by -32 (reverse inequalities): 1.5 ≥ t ≥ 0.5
Final solution: t ∈ (0.5, 1) seconds
At t=0.5 sec, V=80-16=64; at t=1 sec, V=80-32=48 (both within 32-64 range).
📊 Problem Breakdown
- Initial velocity (t=0): V = 80 units/sec
- Deceleration: 32 units/sec2
- Time to zero velocity: t = 80/32 = 2.5 sec
- Target velocity range: 32 to 64 units/sec
❌ Why Other Options Fail
(A) (1, 3/2): At t=1, V=48 (OK), but t=1.5 gives V=32 exactly—velocity drops below 32 afterward.
(C) (1/2, 3/2): Spans 0.5-1.5 sec; at t=1.5, V=32, but exceeds the precise interval where V stays strictly between 32-64.
(D) (1, 3): t=1 V=48 (OK), but t=2 V=16<32, and t=2.5 V=-10 (downward motion).
🔬 Key Physics Insight
Upward velocity decreases linearly from 80 to 0 over 2.5 seconds. Use compound inequalities for kinematics problems in exams like GATE. The interval (1/2, 1) precisely captures when velocity remains between 32-64 units/sec during upward motion.
